Difficult Chain Rule Problem

**funnylookinkid** · March 30th 2008, 08:41 PM

I believe it uses the chain rule, anyway. My professor told me in class the other day that the chain rule can be used so I would prefer if there was an answer utilizing the rule, but really, anything will do. These webwork problems kill me! TIA!

If

,

Calculate

.

help!

**Jhevon** · March 30th 2008, 09:21 PM

Originally Posted by funnylookinkid

I believe it uses the chain rule, anyway. My professor told me in class the other day that the chain rule can be used so I would prefer if there was an answer utilizing the rule, but really, anything will do. These webwork problems kill me! TIA!

If

,

Calculate

.

help!

By the chain rule (look this up): $\frac d{dx} (f(3x^2)) = f'(3x^2) \cdot 6x$

we are told that this is equal to $7x^4$

thus we have $f'(3x^2) \cdot 6x = 7x^4$

$\Rightarrow f'(3x^2) = \frac 76x^3$

i leave the rest to you

**funnylookinkid** · March 30th 2008, 09:38 PM

I get all of that and I get that dy/dx = dy/du * du/dx and I have what you have above. I just cannot figure out how to relate the 3x^2 to x and what to do to differentiate at x. I'm trying to use all of the classic formulas of the chain rule but I have found myself quite baffled! I'm not fishing for an answer here, but maybe there's something I'm missing?

**Jhevon** · March 30th 2008, 09:41 PM

Originally Posted by funnylookinkid

I get all of that and I get that dy/dx = dy/du * du/dx and I have what you have above. I just cannot figure out how to relate the 3x^2 to x and what to do to differentiate at x. I'm trying to use all of the classic formulas of the chain rule but I have found myself quite baffled! I'm not fishing for an answer here, but maybe there's something I'm missing?

there is no more calculus left to be done. the rest is basically using your knowledge of composite functions. what does $f'(3x^2)$ mean? this is a composite function. it means you take $f'(x)$ and replace all the x's you see in it with $3x^2$ . just work backwards.

**funnylookinkid** · March 30th 2008, 09:45 PM

I arrived at (7/18)x and it is not registering as a correct answer. Is that what I should have arrived at?

**funnylookinkid** · March 30th 2008, 09:50 PM

I figured since you were plugging in 3x^2 for all xs, I set up the following:

3x^2 * (t) = (7/6)x^3, where t = what I had to multiply 3x^2 by in order to reach my final answer. The answer I got was (7/18)x. I guess I'm missing something?

**Jhevon** · March 30th 2008, 09:55 PM

Originally Posted by funnylookinkid

I figured since you were plugging in 3x^2 for all xs, I set up the following:

3x^2 * (t) = (7/6)x^3, where t = what I had to multiply 3x^2 by in order to reach my final answer. The answer I got was (7/18)x. I guess I'm missing something?

ok, so the reason you are having trouble with this is that you do not have a good grasp of composite functions. making composite function and multiplying functions are two totally different things! replacing x with a function is not the same as multiplying by the function.

if you have $f(x) = x^2$

then $f(x + 2) = (x + 2)^2$ NOT $x^2(x + 2)$

now try to work from that idea. you want to somehow write the function in terms of $3x^2$ , and then replace the $3x^2$ with $x$ . that is all

**funnylookinkid** · March 30th 2008, 10:07 PM

Okay. I think I understand now. Here's what I did now:

I rewrote the function of (7/6)x^3 as 3x^2 * (7/18)x, since I'm trying to rewrite the derivative in terms of 3x^2 so I can replace it with just x. I did that and was left with x * (7/18)x, multiplied it out to get (7/18)x^2.

But the answer is still wrong. I thought I had a pretty good grasp on this stuff, but apparently I was wrong! Am I still missing something?

**Jhevon** · March 30th 2008, 10:10 PM

Originally Posted by funnylookinkid

Okay. I think I understand now. Here's what I did now:

I rewrote the function of (7/6)x^3 as 3x^2 * (7/18)x, since I'm trying to rewrite the derivative in terms of 3x^2 so I can replace it with just x. I did that and was left with x * (7/18)x, multiplied it out to get (7/18)x^2.

But the answer is still wrong. I thought I had a pretty good grasp on this stuff, but apparently I was wrong! Am I still missing something?

that is wrong. you have an x left over. the only variable you want in your function is 3x^2, you want no x's beside it or anything. it must only be 3x^2 times some constant. try once more (plus, you should have realized that if you plug in 3x^2 into (7/18)x^2 you do not get (7/6)x^3 as you should)

**funnylookinkid** · March 30th 2008, 10:33 PM

I think I'm gonna have to sleep on it, haha. I'll be back in the morning, undoubtedly, hopefully with something more than I have now!

**Jhevon** · March 30th 2008, 10:35 PM

Originally Posted by funnylookinkid

I think I'm gonna have to sleep on it, haha. I'll be back in the morning, undoubtedly, hopefully with something more than I have now!

note that $\frac 76x^3 = \frac 76 \left(\frac {{\color{red}3x^2}}3 \right)^{3/2}$

how did i get that?

i'm leaving, good luck

**DivideBy0** · March 30th 2008, 10:46 PM

Originally Posted by Jhevon

By the chain rule (look this up): $\frac d{dx} (f(3x^2)) = f'(3x^2) \cdot 6x$

we are told that this is equal to $7x^4$

thus we have $f'(3x^2) \cdot 6x = 7x^4$

$\Rightarrow f'(3x^2) = \frac 76x^3$

i leave the rest to you

I like this guy's method, watch and learn

YouTube - 1991 AHSME #21 - Functions

**ThePerfectHacker** · March 31st 2008, 07:17 AM

Originally Posted by DivideBy0

I like this guy's method, watch and learn

YouTube - 1991 AHSME #21 - Functions

I seen other stuff by this guy. The only problem with his videos is he talks way too fast. But that is probably because he needs to explain all of that in under 10 minute, for that is the boundary.

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